Let $a(x)$ and $b(x)$ be smooth functions, i.e they are infinitely times differentiable.

I have made the assumption that the derivative for the function

$$f(x)= (a\cdot b)(x)$$

can be given by

$$f^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}a^{(k)}b^{(n-k)}(x)$$

Where I have defined $g^{(n)}$ as the $n-$ th derivative of some function.

I have made the observation when I was trying to calculate the derivative of such a function and I noticed that the derivatives are in the form:

$$f^{(1)}(x)=a'b+b'a$$

$$f^{(2)}(x)=ab''+2a'b'+a''b$$

$$f^{(3)}(x)=ab'''+3a'b''+3a''b'+a'''b$$

What I have tried so far is induction but I don't know how to manipulate the formula to get the result I want

$${f^{(n+1)}=f^{(n)}}^{'}=(\sum_{k=0}^{n}\binom{n}{k}a^{(k)}b^{(n-k)})^{'}=(\sum_{k=0}^{n}\binom{n}{k}[a^{(k+1)}b^{(n-k)}+a^{(k)}b^{(n-k+1)}])$$

Now I **cannot** say something like :

$$[\uparrow]=(\sum_{k=0}^{n}\binom{n}{k}[a^{(k)}b^{(n-k)}(a+b)]) $$

Because they are derivatives and they do not necessarily behave like exponentials. But I don't know if it even would help me

Because in the end I want to have a term that Looks like

$$\sum_{k=0}^{n+1}\binom{n+1}{k}a^kb^{n-k}$$

But I don't know how to do it.